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## Modulation

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**In general, for DSB-SC, we have**X m(t) xc(t) coswct**A general sketch of the spectrum is shown below.**Xc(f) J0(kp) J1(kp) J-1(kp) J-2(kp) J2(kp) f fc-2fm fc-fm fc fc+fm fc+2fm**Instead of**we will have**The reason for this change can be seen by finding the**corresponding instantaneous frequency: We see that the frequency varies about a nominal value wc by kfm(t).**The quantity kf/wm is called the modulation index and is**denoted by the letter b.**The power spectral density of the noise is the power per**Hzof the noise. Dimensionally, Power = (Power Spectral Density) x (Bandwidth). In general, when the power spectral density is not constant, we can find the power from the power spectral density using**Parseval’s Theorem for Fourier Series**Given a signal described by a Fourier series, the power can be found from its Fourier series coefficients by using something called Parseval’s Theorem for Fourier series.**It can be shown that (exercise for the student):**This function acts like a selector function sifting out those terms for which n=m. So,**Parseval’s Theorem for Fourier Transforms**Parseval’s theorem for Fourier transforms uses Fourier transformsto calculate the energy in a signal. Energy is equal to power times time. If the power is not constant, energy is equal to the integral of the power over time.**The Quadrature Decomposition of Noise**Very often noise is added to a modulated signal, that is a signal whose frequency components are centered about some high frequency fc. The corresponding noise is usually passed through a band-pass filter in the demodulation process. The noise of interest is a band-pass signal.**A bandpass signal can be represented as a modulated form of**lowpass signals. The lowpass signals are, effectively, upconvertedto bandpass signals. If we have a bandpass noise signal n(t), it can be represented in terms of lowpass signals nc(t)and ns(t):**The power in each the in-phase and quadrature**components,nc(t), ns(t), are the same as that of the noise itself, n(t).**The power spectral density of n(t) is proportional to the**square of its spectrum. Sn(f) A2/2 f B The power in n(t) can be calculated by integrating the power spectral density.**P = 2 [ B•A2/2] = BA2.**If we compute the power spectral density and the power of nc(t)and ns(t) we get the same value: Snc(f), Sns(f) A2 f B**The calculation of the signal-to-noise improvement (SNRI) is**shown in the following diagram. so(t) si(t) Demodulator ni(t) no(t)**Let s(t) be a noiseless digital signal (the transmitted**signal). Let n(t) be additive Gaussian noise. The signal n(t) is a random process as a function of time. At each instant in time, n(t) is equal to a is a random variable n which has a Gaussian distribution. The received signal (with noise) is**The noise random variable n has a Gaussian probability**density function with a variance of s2. [We have (n/s) in place of n and an extra s term in the denominator.]**Example: Suppose we were transmitting TTL with Gaussian**noise whose variance is unity. Plot the distribution functions for the received signal. Solution:The probability density functions would be**The probability that the noise is greater than 2.5 volts can**be found by integrating the probability density function for the noise.**We can formally solve this unequal logic level likelihood**problem by starting with the following: In words, if the probability of a logic one is greater than the probability of a logic zero (given received voltage values, logic level likelihoods, et al.), then we decide in favor of a logic zero. Likewise for logic one.**Inserting these expressions into our original decision rule,**we have Or,**To find-out the BER, we use the theorem of total**probability:**1**0 1 1 0 1**To generate an OOK signal, we simply multiply the digital**(baseband) signal by the unmodulated carrier. X**In a variation of OOK, we multiply the sinewave by a bipolar**or antipodal version of the digital waveform: 1 0 1 1 0 1 + + + + 0 volts - - - -**The resultant modulated waveform is actually a form of**phase-modulation called BPSK. X**Since we have four phases, we cannot simply assign these**phases to logic one and logic zero. Instead, we assign each of the four phases to pairs of bits. 01 11 00 10**digitally-modulated**signal X We must also interpret the output of the integrator appropriately. The value of the output will determine if input signal corresponds to a one or a zero.**ds(t)**X s(t) s cos wct The products(t) cos wct is denoted by ds(t) We denote the output of the integrator by s.**Now, if**what would ds(t) and s be?